Question

y=2cos方x+2根号3sinxcosx的最小正周期得多少,

Answer 1

y=2(1+cos2x)/2+√3(2sinxcosx)

=√3sin2x+cos2x+1

=2(sin2x*√3/2+cos2x*1/2)+1

=2(sin2xcosπ/6+cos2xsinπ/6)+1

=2sin(2x+π/6)+1

所以T=2π/2=π